More R/W on a dry or moist day?

[TP4]

Wondering if anyone can shed light on the issue of whether an aircraft will have a longer ground roll on a day that is dry (ie low relative humidity), or on a moist day (ie high relative humidity) and why??

[Elroy Jettson]

At 60 degrees f, jumping from zero to 50% humidity will cost you roughly 1% in air density. You might like to think about it like this. For a given volume, the water molecules displace the air molecules.

Therefore from lift formula, CL 1/2 rho Vsquared S, you need a greater V or TAS as in this case. Greater TAS, higher groundspeed, more runway used to achieve the same IAS for takeoff.

Also ground run increases because less thrust or HP for props, produced by the engine, not taking into account flat rated engines below their rated temp. Hope this helps.

A very simplified answer, I await the boffins inevitable expansion.

[Midland63]

You might like to think about it like this. For a given volume, the water molecules displace the air molecules.

That is a helpful way of thinking about but (whilst not doubting what you say) explain to me why the water molecules don't do the same job as the air molecules?

Thanks

[Elroy Jettson]

Midland 63, good question, but there is a rather simple reason why.

Most people find it hard to believe that humid air is lighter, or less dense, than dry air. How can the air become lighter if we add water vapour to it?
Scientists have known this for a long time. The first was Isaac Newton, who stated that humid air is less dense than dry air.

To see why humid air is less dense than dry air, we need to turn to one of the laws of nature the Italian physicist Amadeo Avogadro discovered in the early 1800s. In simple terms, he found that a fixed volume of gas, say one cubic meter, at the same temperature and pressure, would always have the same number of molecules no matter what gas is in the container.

Imagine a cubic meter of perfectly dry air. It contains about 78% nitrogen molecules, which each have a molecular weight of 28 (2 atoms with atomic weight 14) . Another 21% of the air is oxygen, with each molecule having a molecular weight of 32 (2 stoms with atomic weight 16). The final one percent is a mixture of other gases, which we won't worry about.
Molecules are free to move in and out of our cubic meter of air. What Avogadro discovered leads us to conclude that if we added water vapor molecules to our cubic meter of air, some of the nitrogen and oxygen molecules would leave — remember, the total number of molecules in our cubic meter of air stays the same.

The water molecules, which replace nitrogen or oxygen, have a molecular weight of 18. (One oxygen atom with atomic weight of 16, and two hydrogen atoms each with atomic weight of 1). This is lighter than both nitrogen and oxygen. In other words, replacing nitrogen and oxygen with water vapor decreases the weight of the air in the cubic foot; that is, it's density decreases.

You might say, "I know water's heavier than air." True, liquid water is heavier, or more dense, than air. But, the water that makes the air humid isn't liquid. It's water vapor, which is a gas that is lighter than nitrogen or oxygen.

So, in effect we know from Bernoulis theory, "In the streamlined flow of an ideal fluid, the sum of the pressures are constant", in this case, the fluid is the air moving over the surface of the wing. So we have reduced the density of the air, therefore we have reduced the pressure it exerts, that is the lift on the wing.

Same with the engine. The air is less dense, produces less work.

Hope this answers it for you.

Cheers. :)

[Midland63]

Thanks Elroy

Or in even simpler terms, the water molecules don't do the same job as the air (O2 & N2) molecules they've booted out the way in accordance with Avogadro's law because they (the H2O molecules) are lighter, right?

[Elroy Jettson]

Yep.

Or in even simpler terms, "cos they just bl@@dy do!" :}

[chornedsnorkack]

Does the different adiabatic behaviour of the water vapour also affect lift?

A wing flies because it generates increased pressure underneath as well as decreased pressure above.

If the wing lower surface meets the air, the air is compressed. Adiabatic compression would mean that the air underneath the wing must be warmer than the free air ahead.

Now water vapour has greater heat capacity than nitrogen and oxygen - it has an extra degree of freedom for rotation. Plus more vibrations allowed.

So, moist air does not warm as much as dry air when compressed - and therefore does not create as much extra pressure.

Analogically, moist air cooling above the wing should have higher temperature and higher pressure than dry air.

Is this a significant issue?

[barit1]

Humidity affects engine performance in a similar way (i.e. reduced air density and thus mass flow) but there are additional complex effects on each component; specific heat (which changes the internal pressure-to-temperature relationships), and inlet condensation (which heats the inlet air as static pressure drops).

IIRC the published data are based on a fairly high standard humidity level (85%??) so if humidity is low, your performance will be better than book.

OTOH, at very high relative humidity, it's prudent to give yourself a little extra margin.

[TP4]

Elroy Jettson,

Very insightful. Cleared that issue up for me.

Thanks.:ok:

[arismount]

Dry air is composed of approx. 78% of diatomic nitrogen (N2 molecule) with a molecular weight of 14 (Atomic weight of N=7), and approx. 21% of diatomic oxygen (O2 molecule) with a molecular weight of 16 (Atomic weight of O=8).

If the air is humid, i.e., contains water vapor, the vapor molecules have a molecular weight of 10 (H2O molecular weight of 10 = 2 atom of Hydrogen, atomic weight 1, and one atom of oxygen, atomic weight 8).

From this you can see that a "particle" of water vapor is less dense than either a "particle" of nitrogen or oxygen.

OK. Let's see if we could come up with a "molecular weight" for dry air, i.e., treat the mixture that is air as if it were a compound. For simplicity, let's neglect the other gases that are in air, and also proportionately round up the percentages of N & O so that the total equals 100%:

Then: 78.75% N + 21.25% O = 100% Air.
(.7875 x 14) + (.2125 x 16) = 14.43. That is, if dry air were a chemical compound instead of a mixture, said compound would have a molecular weight of 14.43.

Therefore, neglecting specific measurement systems, we could say that a given parcel of air weighed 14.43 "units."

Now let's see what injecting some water vapor into the dry air would do.

John Dalton's law of partial pressures states that the Total Pressure of a gas mixture was the sum of the Partial Pressure of each gas. That is, for a given atmospheric pressure, that pressure is the sum total of the pressures of each gas in the mixture of gases that is air. In the case of humid air, one such gas in the mixture is water vapor.

OK, let's hypothesize that a parcel of humid air contains 3% water vapor.

(.97 x 14.43) + (.03 x 10) = 14.29. That is, a parcel of such air would weigh onl 14.29 units, a decrease of 0.14 units, or 1%.

Since the lift developed is proportional to the density of the air, this humid parcel of air would develop 1% less lift than dry air.

[AirRabbit]

...the water molecules displace the air molecules.
Dang! I always thought that water vapor displaced lift. The more water vapor, the less lift. Eventually, you get to the point where there just isn’t enough lift to fly. That is why smart pilots and birds don’t attempt to fly when there is too much water vapor in the air.